oj <- OJ
Create a training set containing a random sample of 800 observations, and a test set containing the remaining observations.
set.seed(1237)
train <- sample(1:nrow(oj), 800)
oj.train <- oj[train,]
oj.test <- oj[-train,]
nrow(oj.train)
## [1] 800
nrow(oj.test)
## [1] 270
Fit a tree to the training data, with Purchase as the response and the other variables as predictors. Use the summary() function to produce summary statistics about the tree, and describe the results obtained. What is the training error rate? How many terminal nodes does the tree have?
What variables are predictors:
names(oj)
## [1] "Purchase" "WeekofPurchase" "StoreID" "PriceCH"
## [5] "PriceMM" "DiscCH" "DiscMM" "SpecialCH"
## [9] "SpecialMM" "LoyalCH" "SalePriceMM" "SalePriceCH"
## [13] "PriceDiff" "Store7" "PctDiscMM" "PctDiscCH"
## [17] "ListPriceDiff" "STORE"
Creating the tree:
oj.rpart0 <- rpart(Purchase ~ WeekofPurchase + StoreID + PriceCH + PriceMM + DiscCH + DiscMM + SpecialCH + SpecialMM + LoyalCH + SalePriceMM + SalePriceCH + PriceDiff + Store7 + PctDiscMM + PctDiscCH + ListPriceDiff + STORE,
data = oj.train)
printcp(oj.rpart0) # display the results
##
## Classification tree:
## rpart(formula = Purchase ~ WeekofPurchase + StoreID + PriceCH +
## PriceMM + DiscCH + DiscMM + SpecialCH + SpecialMM + LoyalCH +
## SalePriceMM + SalePriceCH + PriceDiff + Store7 + PctDiscMM +
## PctDiscCH + ListPriceDiff + STORE, data = oj.train)
##
## Variables actually used in tree construction:
## [1] ListPriceDiff LoyalCH PctDiscCH
##
## Root node error: 308/800 = 0.385
##
## n= 800
##
## CP nsplit rel error xerror xstd
## 1 0.519481 0 1.00000 1.00000 0.044685
## 2 0.016234 1 0.48052 0.49351 0.036026
## 3 0.010000 4 0.42208 0.48377 0.035751
# summary(oj.rpart0) # detailed summary of splits
# plotcp(oj.rpart0) # visualize cross-validation results
# multiple ways to calculate the training error rate:
class.pred <- table(predict(oj.rpart0, type="class"), oj.train$Purchase)
1-sum(diag(class.pred))/sum(class.pred)
## [1] 0.1625
0.385*0.42208
## [1] 0.1625008
We can see that the training error rate is 16.25%, and 5 nodes from 4 splits.
Type in the name of the tree object in order to get a detailed text output. Pick one of the terminal nodes, and interpret the information displayed.
oj.rpart0
## n= 800
##
## node), split, n, loss, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 308 CH (0.61500000 0.38500000)
## 2) LoyalCH>=0.450956 514 85 CH (0.83463035 0.16536965)
## 4) LoyalCH>=0.705699 296 16 CH (0.94594595 0.05405405) *
## 5) LoyalCH< 0.705699 218 69 CH (0.68348624 0.31651376)
## 10) ListPriceDiff>=0.235 134 22 CH (0.83582090 0.16417910) *
## 11) ListPriceDiff< 0.235 84 37 MM (0.44047619 0.55952381)
## 22) PctDiscCH>=0.052007 12 2 CH (0.83333333 0.16666667) *
## 23) PctDiscCH< 0.052007 72 27 MM (0.37500000 0.62500000) *
## 3) LoyalCH< 0.450956 286 63 MM (0.22027972 0.77972028) *
# summary(oj.rpart0)
We can see from this output that at terminal node #3 the observations are tested for LoyalCH< 0.450956; 286 observations from the training dataset end up at this node and 63 are misclassified.
Create a plot of the tree, and interpret the results.
Borrowing some code from the script provided in class:
plot(oj.rpart0, uniform=TRUE, main="Regression Tree for Purchase ")
text(oj.rpart0, use.n=TRUE, all=TRUE, cex=.8)
This plot shows more clearly what is tabulated in the results from part c; namely that there are 5 nodes, and four splits.
Predict the response on the test data, and produce a confusion matrix comparing the test labels to the predicted test labels. What is the test error rate?
pred.oj.rpart0 <- predict(oj.rpart0,
newdata = oj.test,
type = "class")
confusionMatrix(pred.oj.rpart0, oj.test$Purchase)
## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 132 24
## MM 29 85
##
## Accuracy : 0.8037
## 95% CI : (0.7512, 0.8494)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : 2.768e-13
##
## Kappa : 0.5953
##
## Mcnemar's Test P-Value : 0.5827
##
## Sensitivity : 0.8199
## Specificity : 0.7798
## Pos Pred Value : 0.8462
## Neg Pred Value : 0.7456
## Prevalence : 0.5963
## Detection Rate : 0.4889
## Detection Prevalence : 0.5778
## Balanced Accuracy : 0.7998
##
## 'Positive' Class : CH
##
1-0.8037
## [1] 0.1963
The test error rate is \(1-0.8037 = 0.1963\).
Apply the cv.tree() function to the training set in order to determine the optimal tree size.
We will use prune() instead of cv.tree, to determine the optima tree size:
# prune the tree for min CV error
pfit<- prune(oj.rpart0,
cp = oj.rpart0$cptable[which.min(oj.rpart0$cptable[,"xerror"]),"CP"])
pfit
## n= 800
##
## node), split, n, loss, yval, (yprob)
## * denotes terminal node
##
## 1) root 800 308 CH (0.61500000 0.38500000)
## 2) LoyalCH>=0.450956 514 85 CH (0.83463035 0.16536965)
## 4) LoyalCH>=0.705699 296 16 CH (0.94594595 0.05405405) *
## 5) LoyalCH< 0.705699 218 69 CH (0.68348624 0.31651376)
## 10) ListPriceDiff>=0.235 134 22 CH (0.83582090 0.16417910) *
## 11) ListPriceDiff< 0.235 84 37 MM (0.44047619 0.55952381)
## 22) PctDiscCH>=0.052007 12 2 CH (0.83333333 0.16666667) *
## 23) PctDiscCH< 0.052007 72 27 MM (0.37500000 0.62500000) *
## 3) LoyalCH< 0.450956 286 63 MM (0.22027972 0.77972028) *
Produce a plot with tree size on the x-axis and cross-validated classification error rate on the y-axis.
par(mfrow=c(1,3)) # three plots on one page
#This produces the first plot, asked for in part g
plotcp(pfit)
# This produces the other plots of number of splits vs R2 & X-relative error
rsq.rpart(pfit) # visualize cross-validation results
##
## Classification tree:
## rpart(formula = Purchase ~ WeekofPurchase + StoreID + PriceCH +
## PriceMM + DiscCH + DiscMM + SpecialCH + SpecialMM + LoyalCH +
## SalePriceMM + SalePriceCH + PriceDiff + Store7 + PctDiscMM +
## PctDiscCH + ListPriceDiff + STORE, data = oj.train)
##
## Variables actually used in tree construction:
## [1] ListPriceDiff LoyalCH PctDiscCH
##
## Root node error: 308/800 = 0.385
##
## n= 800
##
## CP nsplit rel error xerror xstd
## 1 0.519481 0 1.00000 1.00000 0.044685
## 2 0.016234 1 0.48052 0.49351 0.036026
## 3 0.010000 4 0.42208 0.48377 0.035751
## Warning in rsq.rpart(pfit): may not be applicable for this method
Which tree size corresponds to the lowest cross-validated classification error rate?
As we can see from the first plot above a tree size of 5 (i.e. 4 splits) results in the lowest CV error rate
Produce a pruned tree corresponding to the optimal tree size obtained using cross-validation. If cross-validation does not lead to selection of a pruned tree, then create a pruned tree with five terminal nodes.
As we can see from the above results, the tree produced by the prune() function has 5 terminal nodes and has the optimal cross-validated error.
Compare the training error rates between the pruned and unpruned trees. Which is higher?
par(mfrow=c(1,3)) # three plots on one page
plot(oj.rpart0, uniform=TRUE, main="Regression Tree for Purchase ")
text(oj.rpart0, use.n=TRUE, all=TRUE, cex=.8)
plot(pfit, uniform=TRUE, main="PrunedRegression Tree for Purchase ")
text(pfit, use.n=TRUE, all=TRUE, cex=.8)
As we can see from the above plot, the pruned and unpruned trees are identical. We can calculate their training error rates:
#unpruned training error rate:
class.pred <- table(predict(oj.rpart0, type="class"), oj.train$Purchase)
1-sum(diag(class.pred))/sum(class.pred)
## [1] 0.1625
#pruned training error rate:
class.pred <- table(predict(pfit, type="class"), oj.train$Purchase)
1-sum(diag(class.pred))/sum(class.pred)
## [1] 0.1625
As expected, they are identical.
Compare the test error rates between the pruned and unpruned trees. Which is higher?
pred.pfit <- predict(pfit,
newdata = oj.test,
type = "class")
confusionMatrix(pred.oj.rpart0, oj.test$Purchase)
## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 132 24
## MM 29 85
##
## Accuracy : 0.8037
## 95% CI : (0.7512, 0.8494)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : 2.768e-13
##
## Kappa : 0.5953
##
## Mcnemar's Test P-Value : 0.5827
##
## Sensitivity : 0.8199
## Specificity : 0.7798
## Pos Pred Value : 0.8462
## Neg Pred Value : 0.7456
## Prevalence : 0.5963
## Detection Rate : 0.4889
## Detection Prevalence : 0.5778
## Balanced Accuracy : 0.7998
##
## 'Positive' Class : CH
##
confusionMatrix(pred.pfit, oj.test$Purchase)
## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 132 24
## MM 29 85
##
## Accuracy : 0.8037
## 95% CI : (0.7512, 0.8494)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : 2.768e-13
##
## Kappa : 0.5953
##
## Mcnemar's Test P-Value : 0.5827
##
## Sensitivity : 0.8199
## Specificity : 0.7798
## Pos Pred Value : 0.8462
## Neg Pred Value : 0.7456
## Prevalence : 0.5963
## Detection Rate : 0.4889
## Detection Prevalence : 0.5778
## Balanced Accuracy : 0.7998
##
## 'Positive' Class : CH
##
1-0.8037
## [1] 0.1963
As expected test error rates are identical.
Apply random forest (of your choice) and boosting (of your choice) for the data set in #9 and compare the results (test errors). Which one performs best?
set.seed(1237)
# Create Random forest
oj.cforest0 <- cforest(Purchase ~ .,
data = oj.train,
control = cforest_unbiased(ntree = 500))
# Predict test results:
pcforest0 <- predict(oj.cforest0,
newdata = oj.test,
OOB = TRUE,
type = "response")
# Confusion Matrix:
confusionMatrix(pcforest0, oj.test$Purchase)
## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 144 32
## MM 17 77
##
## Accuracy : 0.8185
## 95% CI : (0.7673, 0.8626)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : 3.798e-15
##
## Kappa : 0.6145
##
## Mcnemar's Test P-Value : 0.0455
##
## Sensitivity : 0.8944
## Specificity : 0.7064
## Pos Pred Value : 0.8182
## Neg Pred Value : 0.8191
## Prevalence : 0.5963
## Detection Rate : 0.5333
## Detection Prevalence : 0.6519
## Balanced Accuracy : 0.8004
##
## 'Positive' Class : CH
##
Preparing the datasets:
oj2 <- oj
oj2$Store7 <- as.numeric(oj2$Store7)
oj2.train <- oj2[train,]
oj2.test <- oj2[-train,]
Gradient boosting:
set.seed(1237)
# Creating OJ Purchase as the y-variable:
y <- oj2.train$Purchase
num.class = length(levels(y))
y <- as.numeric(y)-1 #y should start from 0 and numeric, not factor.
# Construct xgb.DMatrix object
dtrain <- xgb.DMatrix(as.matrix(oj2.train[,2:18]), label = y)
param <- list("objective" = "multi:softprob","num_class" = 2)
#param <- list("objective" = "multi:softprob",
# "num_class" = 3,
# "num_parallel_tree" = 1,
# "eval_metric" = "mlogloss",
# "nthread" = 8,
# "max_depth" = 6,
# "eta" = 0.3, #learning rate
# "gamma" = 0,
# "subsample" = 1, #subsample ratio
# "colsample_bytree" = 1, #subsample ratio of columns
# "min_child_weight" = 1)
xfit1 <- xgboost(param=param,
dtrain,
nrounds=3) #compare with nround =2, num_parallel_tree seems to be 3 (?)
## [22:55:36] WARNING: amalgamation/../src/learner.cc:1115: Starting in XGBoost 1.3.0, the default evaluation metric used with the objective 'multi:softprob' was changed from 'merror' to 'mlogloss'. Explicitly set eval_metric if you'd like to restore the old behavior.
## [1] train-mlogloss:0.535375
## [2] train-mlogloss:0.445079
## [3] train-mlogloss:0.382117
print(xfit1)
## ##### xgb.Booster
## raw: 28.4 Kb
## call:
## xgb.train(params = params, data = dtrain, nrounds = nrounds,
## watchlist = watchlist, verbose = verbose, print_every_n = print_every_n,
## early_stopping_rounds = early_stopping_rounds, maximize = maximize,
## save_period = save_period, save_name = save_name, xgb_model = xgb_model,
## callbacks = callbacks)
## params (as set within xgb.train):
## objective = "multi:softprob", num_class = "2", validate_parameters = "TRUE"
## xgb.attributes:
## niter
## callbacks:
## cb.print.evaluation(period = print_every_n)
## cb.evaluation.log()
## # of features: 17
## niter: 3
## nfeatures : 17
## evaluation_log:
## iter train_mlogloss
## 1 0.535375
## 2 0.445079
## 3 0.382117
xgb.plot.tree(model = xfit1)
xgb.plot.multi.trees(model = xfit1)
## Column 2 ['No'] of item 2 is missing in item 1. Use fill=TRUE to fill with NA (NULL for list columns), or use.names=FALSE to ignore column names. use.names='check' (default from v1.12.2) emits this message and proceeds as if use.names=FALSE for backwards compatibility. See news item 5 in v1.12.2 for options to control this message.
#xgb.dump(xfit1, with_stats = T)
pred = predict(xfit1,dtrain)
pred = matrix(pred,ncol=3,byrow=T)
## Warning in matrix(pred, ncol = 3, byrow = T): data length [1600] is not a sub-
## multiple or multiple of the number of rows [534]
pclass = apply(pred,1,which.max)
# table(oj2.train$Purchase,pclass)
# confusionMatrix(unique(oj2.train$Purchase)[pclass],oj2.train$Purchase)
mat = xgb.importance(feature_names = colnames(oj[,-5]),model = xfit1)
xgb.plot.importance(importance_matrix = mat)
Creating testing set confusion matrix:
# Creating OJ Purchase as the y-variable:
y2 <- oj2.test$Purchase
y2 <- as.numeric(y2)-1 #y should start from 0 and numeric, not factor.
dtest <- xgb.DMatrix(as.matrix(oj2.test[,2:18]), label = y2)
pred.test <- predict(xfit1,
newdata = dtest,
strict_shape = TRUE)
pred_labels <- factor(round(pred.test[2,])+1, labels = c("CH", "MM"))
# pred_labels <- max.col(pred.test) - 1
test_labels <- as.factor(oj2.test$Purchase)
confusionMatrix(pred_labels, oj2.test$Purchase)
## Confusion Matrix and Statistics
##
## Reference
## Prediction CH MM
## CH 141 25
## MM 20 84
##
## Accuracy : 0.8333
## 95% CI : (0.7834, 0.8758)
## No Information Rate : 0.5963
## P-Value [Acc > NIR] : <2e-16
##
## Kappa : 0.6512
##
## Mcnemar's Test P-Value : 0.551
##
## Sensitivity : 0.8758
## Specificity : 0.7706
## Pos Pred Value : 0.8494
## Neg Pred Value : 0.8077
## Prevalence : 0.5963
## Detection Rate : 0.5222
## Detection Prevalence : 0.6148
## Balanced Accuracy : 0.8232
##
## 'Positive' Class : CH
##
The XGBoost method produces the best test set accuracy, with accuracy of 83.3%.
sessionInfo()
## R version 4.1.2 (2021-11-01)
## Platform: x86_64-w64-mingw32/x64 (64-bit)
## Running under: Windows 10 x64 (build 19044)
##
## Matrix products: default
##
## locale:
## [1] LC_COLLATE=English_United States.1252
## [2] LC_CTYPE=English_United States.1252
## [3] LC_MONETARY=English_United States.1252
## [4] LC_NUMERIC=C
## [5] LC_TIME=English_United States.1252
##
## attached base packages:
## [1] stats4 grid stats graphics grDevices utils datasets
## [8] methods base
##
## other attached packages:
## [1] DiagrammeR_1.0.8 xgboost_1.5.0.2 party_1.3-9 strucchange_1.5-2
## [5] sandwich_3.0-1 zoo_1.8-9 modeltools_0.2-23 mvtnorm_1.1-3
## [9] randomForest_4.7-1 rattle_5.4.0 bitops_1.0-7 tibble_3.1.6
## [13] rpart_4.1.16 caret_6.0-90 lattice_0.20-45 ISLR2_1.3-1
## [17] ggplot2_3.3.5
##
## loaded via a namespace (and not attached):
## [1] nlme_3.1-155 matrixStats_0.61.0 lubridate_1.8.0
## [4] RColorBrewer_1.1-2 tools_4.1.2 utf8_1.2.2
## [7] R6_2.5.1 DBI_1.1.2 colorspace_2.0-2
## [10] nnet_7.3-17 withr_2.4.3 tidyselect_1.1.1
## [13] compiler_4.1.2 scales_1.1.1 proxy_0.4-26
## [16] stringr_1.4.0 digest_0.6.29 rmarkdown_2.11
## [19] pkgconfig_2.0.3 htmltools_0.5.2 parallelly_1.30.0
## [22] fastmap_1.1.0 highr_0.9 htmlwidgets_1.5.4
## [25] rlang_0.4.12 rstudioapi_0.13 visNetwork_2.1.0
## [28] jquerylib_0.1.4 generics_0.1.1 jsonlite_1.7.3
## [31] dplyr_1.0.7 ModelMetrics_1.2.2.2 magrittr_2.0.1
## [34] Matrix_1.4-0 Rcpp_1.0.8 munsell_0.5.0
## [37] fansi_1.0.2 lifecycle_1.0.1 stringi_1.7.6
## [40] multcomp_1.4-18 pROC_1.18.0 yaml_2.2.1
## [43] MASS_7.3-55 plyr_1.8.6 recipes_0.1.17
## [46] parallel_4.1.2 listenv_0.8.0 crayon_1.4.2
## [49] splines_4.1.2 knitr_1.37 pillar_1.6.5
## [52] future.apply_1.8.1 reshape2_1.4.4 codetools_0.2-18
## [55] glue_1.6.1 evaluate_0.14 data.table_1.14.2
## [58] vctrs_0.3.8 foreach_1.5.1 gtable_0.3.0
## [61] purrr_0.3.4 tidyr_1.1.4 future_1.23.0
## [64] assertthat_0.2.1 xfun_0.29 gower_0.2.2
## [67] coin_1.4-2 prodlim_2019.11.13 libcoin_1.0-9
## [70] e1071_1.7-9 class_7.3-20 survival_3.2-13
## [73] timeDate_3043.102 iterators_1.0.13 lava_1.6.10
## [76] globals_0.14.0 TH.data_1.1-0 ellipsis_0.3.2
## [79] ipred_0.9-12